# Cp”gngevtqp.”oqxkpi”cnqpi”vjg”x/czku”ykvj”cp”kpkvkcn”gpgti{ qh”322″gX.”gpvgtu”c”tgikqp”qh”ocipgvke”hkgnf”B ur “?”*307Ù325V+ $k cv”U”*ugg”hkiwtg+0″Vjg”hkgnf”gzvgpfu”dgvyggp”x”?”2″cpf”x”? 4″eo0″Vjg”gngevtqp”ku”fgvgevgf”cv”vjg”rqkpv”S”qp”c”uetggp Question: An electron, moving along the$x$-axis with an initial energy of$100 \mathrm{eV}$, enters a region of magnetic field$\vec{B}=\left(1.5 \times 10^{-3} \mathrm{~T}\right) \hat{k}$at$\mathrm{S}$(see figure). The field extends between$x=0$and$x=2 \mathrm{~cm}$. The electron is detected at the point Q on a screen placed$8 \mathrm{~cm}$away from the point$\mathrm{S}$. The distance$d$between$\mathrm{P}$and$\mathrm{Q}$(on the screen) is : (Electron’s charge$=1.6 \times 10^{-19} \mathrm{C}$, mass of electron$=9.1 \times 10^{-31} \mathrm{~kg}$) 1.$11.65 \mathrm{~cm}$2.$12.87 \mathrm{~cm}$3.$1.22 \mathrm{~cm}$4.$2.25 \mathrm{~cm}$Correct Option: 2 JEE Main Previous Year 1 Question of JEE Main from Physics Moving Charges and Magnetism chapter. JEE Main Previous Year 12 April 2019, II Solution: (2) ### Related Questions • asdf View Solution • questions View Solution • An electron is moving along$+x$direction with a velocity of$6 \times 10^{6} \mathrm{~ms}^{-1}$. It enters a region of uniform electric field of$300 \mathrm{~V} / \mathrm{cm}$pointing along$+y$direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the$x$direction will be: View Solution • A particle of charge$q$and mass$m$is moving with a velocity$-v \hat{i}(v \neq 0)$towards a large screen placed in the$\mathrm{Y}-\mathrm{Z}$plane at a distance$d$. If there is a magnetic field$\vec{B}=B_{0} \hat{k}$, the minimum value of$v$for which the particle will not hit the screen is: View Solution • A charged particle carrying charge$1 \mu \mathrm{C}$is moving with velocity$(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{ms}^{-1}$. If an external magnetic field of$(5 \hat{i}+3 \hat{j}-6 \hat{k}) \times 10^{-3} \mathrm{~T}$exists in the region where the particle is moving then the force on the particle is$\vec{F} \times 10^{-9} \mathrm{~N}$. The vector$\vec{F}$is : View Solution • A beam of protons with speed$4 \times 10^{5} \mathrm{~ms}^{-1}$enters a uniform magnetic field of$0.3 \mathrm{~T}$at an angle of$60^{\circ}$to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton$=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton$=1.69 \times 10^{-19} \mathrm{C}$) View Solution • The figure shows a region of length ‘$l$‘ with a uniform magnetic field of$0.3 \mathrm{~T}$in it and a proton entering the region with velocity$4 \times 10^{5} \mathrm{~ms}^{-1}$making an angle$60^{\circ}$with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘$l$‘ is close to (mass of proton$=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton$=1.6 \times 10^{-19} \mathrm{C}$) View Solution • Proton with kinetic energy of$1 \mathrm{MeV}$moves from south to north. It gets an acceleration of$10^{12} \mathrm{~m} / \mathrm{s}^{2}$by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is$1.6 \times 10^{-27} \mathrm{~kg}$) View Solution • A particle having the same charge as of electron moves in a circular path of radius$0.5 \mathrm{~cm}$under the influence of a magnetic field of$0.5 \mathrm{~T}$. If an electric field of$100 \mathrm{~V} / \mathrm{m}$makes it to move in a straight path then the mass of the particle is (Given charge of electron$=1.6 \times 10^{-19} \mathrm{C}$) View Solution • A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let$\mathrm{r}_{\mathrm{p}}, \mathrm{r}_{\mathrm{e}}$and$\mathrm{r}_{\mathrm{He}}\$ be their respective radii, then,

View Solution

error: Content is protected !!